$\sum\limits_{n=1}^{\infty } \dfrac{(x-3)^n}{n^2}$ What is the interval of convergence of the series? Choose 1 answer: Choose 1 answer: (Choice A) A $-4 < x < -2$ (Choice B) B $-4 \le x \le -2$ (Choice C) C $2\le x \le 4$ (Choice D) D $2 < x < 4$
Explanation: We use the ratio test. For $x\neq3$, let $a_n=\dfrac{(x-3)^n}{n^2}$. $\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|= \left| {x-3}\right| $ The series converges when $\left|{x-3}\right|<1$, which is when $2<x<4$. Now let's check the endpoints, $x=2$ and $x=4$. Letting $x=2$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{(2-3)^n}{n^2} &=\sum\limits_{n=1}^{\infty }\dfrac{(-1)^n}{n^2} \end{aligned}$ By the alternating series test, we know this series converges. Letting $x=4$, we get the series: $\begin{aligned} \dfrac{(4-3)^n}{n^2} &=\sum\limits_{n=1}^{\infty }\dfrac{1}{n^2} \end{aligned}$ By the $p$ -series test, we know this series converges. In conclusion, the interval of convergence is $2\le x \le 4$.